I would like to ask the following question:
Let $G:= \langle x,y \mid \text{some relations} \rangle$.
Is it possible to find a group $G$ as above fulfilling the following criteria simultaneously?
The order $o(x)$ of the element $x$ is finite: $o(x) < \infty$
$o(y) < \infty$
$ N:= \langle x^2 \rangle \unlhd G$
$ |G|=\infty$
$ |G/N| < \infty$
When I tried to find such a group $G$, I always ended up with a finite group $G$, but I would like to have $|G|=\infty$.
Thank you very much for the help.
Since $\lvert x\rvert$ is finite and $N=\langle g= x^2\rangle\unlhd G$ is cyclic, we have
$$g^{\frac{\lvert x\rvert}{2}}=(x^2)^{\frac{\lvert x\rvert}{2}}=x^{\lvert x\rvert}=e,$$
so that $\lvert N\rvert$ is finite. But now
$$\begin{align} \lvert G\rvert&=[G:N]\lvert N\rvert\\ &=\lvert G/N\rvert\lvert N\rvert\\ &<\infty, \end{align}$$
a contradiction.